hello, i am new cfd beginner and i did some simulation of flow around cylinder at various Reynolds in fluent and cfx to get started. I did mesh dependency study and solution is not varying.

when post-processing the global Cd = f(Re) I obtain for both code the well know curve. I have found an example on the web below:

http://imartinez.etsin.upm.es/bk3/c12/Forced%20and%20natural%20convection_archivos/image022.jpg

Now if I also calculate the same local Cd for each nodes or face around the cylinder as a function of local Re (Re(theta)) I obtain the following:

- with cfx every point of each node or edge are located on this same curve (and the integrated average is). - with fluen the points are not on the curve but the integrated average is.

which code is right, fluent or cfx?

should friction versus Re follow this curve both globally AND locally or only the global value.

Later, I plan do look at the Nusselt as well (global and local)

What do you find with the software you use?

thanks in advance.

not many real fluid dynammics specialist on this forum

I'm not sure if I understand your questions, but you will of course not get the same global Cd = f(Re) curve for flow past a cylinder as you will if you for one global Re plot the local Cd = f(Re) for each cell-face on the cylinder.

The global Cd = f(Re) is governed different types of phenomena. At low Re you have a creeping stokes flow where the drag is dependent on friction forces and laminar separation. As the global Re is increased you start getting shear layer instabilities an a von Karman vortex street. At even larger Re, when the boundary layers start to become turbulent and therby avoid separation, you will also see a drag-crisis as a dip on the Cd = f(Re) curve at around 5*10^5. You will of course not see this type of behaviour for the local Cd = f(Re) curve for each cell face on a cylinder at one global Re.

By the way. To siumlate this type of flow, with shear-layer-instabilities, turbulent transition and the drag crisis, is not at all trivial. It is in fact very difficult.

Hello Not a guru, it is not very smart to answer your own question like this.